One mole each of He and $\mathrm{A}(\mathrm{g})$ are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.

$$ \mathrm{A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{~g}) $$

$\mathrm{K}_{\mathrm{c}}$ for this reaction at 400 K is 4.0 . The partial pressures (in atm) of He and B(g) are respectively (at equilibrium)

(Assume $\mathrm{He}, \mathrm{A}(\mathrm{g})$ and $\mathrm{B}(\mathrm{g})$ behave as ideal gases)

(Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

The reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ was initiated with the amount ' a ' of $\mathrm{A}(\mathrm{g})$. At equilibrium it is found that the amount of $\mathrm{A}(\mathrm{g})$ remaining is ( $\mathrm{a}-x$ ) at a total pressure of p .

The equilibrium constant Kp of the reaction can be calculated from the expression :

At $\mathrm{T}(\mathrm{K})$, the equilibrium constant of

$\mathrm{A}_2(g)+\mathrm{B}_2(g) \rightleftharpoons \mathrm{C}(g)$ is $2.7 \times 10^{-5}$.

What is the equilibrium constant for

$\frac{1}{3} \mathrm{~A}_2(\mathrm{~g})+\frac{1}{3} \mathrm{~B}_2(\mathrm{~g}) \rightleftharpoons \frac{1}{3} \mathrm{C}(\mathrm{g})$ at the same temperature?

Observe the following equilibrium in a 1 L flask.

A(g) ⇌ B(g)

At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively